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Question

If x+y+z=0, then ∣∣ ∣∣axbyczcyazbxbzcxay∣∣ ∣∣=xyz∣∣ ∣∣abccabbca∣∣ ∣∣.

A
True
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B
False
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Solution

The correct option is A True
Given,
x+y+z=0 (1)

Taking LHS
∣ ∣axbyczcyazbxbzcxay∣ ∣


=xa(yza2x2bc)yb(y2acb2xz)+zc(c2xyz2ab)=xyz(a3+b3+c3)abc(x3+y3+z3)=xyz(a3+b3+c3)3abc(xyz) .......[since x+y+z=0,t therefore,]

=xyz(a3+b3+c33abc)=xyz(a+b+c)(a2+b2+c2abacbc) (1)

Taking RHS

=xyz∣ ∣abccabbca∣ ∣

Applying R1R1+R2+R3

=xyz∣ ∣a+b+ca+b+ca+b+ccabbca∣ ∣

=xyz(a+b+c)∣ ∣111cabbca∣ ∣

Applying C2C2C1,C3C3C1

=xyz(a+b+c)∣ ∣100cacbcbcbab∣ ∣

=xyz(a+b+c)[(ac)(ab)(cb)(bc)]=xyz(a+b+c)(a2+b2+c2abbcac) (2)


from eqs.(1) and (2)
LHS=RHS

Hence,
∣ ∣axbyczcyazbxbzcxay∣ ∣=xyz∣ ∣abccabbca∣ ∣


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